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Unit 6: Heat and Thermodynamics — Short Questions

11th Class Physics · Unit 6: Heat and Thermodynamics

Exercise Short Questions

6.1.What is meant by thermal equilibrium? Explain briefly.

Thermal Equilibrium

When two bodies are at the same temperature, the thermal energy (which is related to the kinetic energy of particles) of each body is equal. As a result, there is no driving force for heat transfer between them, and thus they remain in thermal equilibrium.

Example

When we put a metal spoon into a hot cup of coffee
(i) initially, the coffee is hotter than the spoon.
(ii) over time, heat flows from the coffee to the spoon
(iii) eventually, the coffee and spoon reach the same temperature

Thermal equilibrium is achieved at this point, there is no net heat flow between the coffee and the spoon, and they are said to be in thermal equilibrium.

6.2.What is meant by internal energy? How is it related to temperature of an idealgas?

Internal Energy
"The sum of all forms of molecular energies (kinetic and potential) of a substance is termed as its internal energy."
The molecules of an ideal gas are mere points masses which exert no force on one another. So, the internal energy of an ideal gas system is generally the translational K.E. of its molecules.

Dependence on temperature Since the temperature of a system is defined as the average K.E. of its molecules. Thus for an ideal gas system, the internal energy is directly proportional to its
temperature. < 1/2 mv² >= 3/2 kB T
Where KB is Boltzmann constant.
Therefore, the rise in temperature of an object represent an increase in internal energy of an ideal gas.

6.3.State 2nd law of thermodynamics in two different forms.

Two common forms of the Second Law of Thermodynamics are:
1. Lord Kelvin Statement:
It is impossible to devise a process which may convert heat taken from a single reservoir entirely into work without leaving any change in the working system.
2. Clausius Statement:
Heat cannot spontaneously flow from a colder body to a hotter body without external work being done on the system.

6.4.Is it possible to construct a heat engine of 100% efficiency? Explain.

No, it's not possible to construct a heat engine with 100% efficiency. According to the Second Law of Thermodynamics, some energy will always be lost as heat, making it impossible to achieve perfect efficiency.
The Carnot efficiency limit (η=1- Tc/Th) suggests a theoretical maximum efficiency, which is always less than 100%. Real-world engines face additional losses, making 100% efficiency unattainable.

6.5.Differentiate between reversible and irreversible processes.

Reversible Process
The process which can be retraced by reversing the controlling factors without producing any change in the surrounding is known as reversible process.

Examples i. Slow expansion and compression of the gas.
ii. Liquefaction and Evaporation.
iii. Melting, freezing and boiling etc.

Irreversible process A process which cannot be retraced in the backward direction by reversing the controlling factors is known as irreversible process.

Example (i) Work done against friction is an irreversible process.
(ii) A chemical explosion.
(iii) All engines in practical life.

6.6.Why adiabat is steeper than isotherm? Explain.

The adiabatic curve is steeper than the isothermal curve because:
In an adiabatic process, no heat is exchanged, so the gas cools faster as it expands, causing pressure to drop more rapidly.
In an isothermal process, temperature remains constant, so pressure drops more gradually during expansion.

6.7.A refrigerator transforms heat from cold to hot body. Does this violate the second law of thermodynamics? Justify your answer.

No, a refrigerator does not violate the second law of thermodynamics. While it transfers heat from a colder body to a hotter one, this is done with the help of external work which is done according to the Clausius statement of the second law.

6.8.Explain briefly heat death of universe in terms of entropy.

The heat death of the universe refers to a theoretical end state where the universe reaches maximum entropy. In this state, all energy is evenly distributed, no temperature differences exist, and no work or useful energy transformations are possible. It means the universe would be in complete thermodynamics equilibrium. Where due to absence of temperature difference no life or processes could occur.

6.9.Is it possible for a cyclic reversible heat engine to absorb heat at constant temperature and transforms it completely into work without rejecting some heat at low temperature? Explain.

No, it is not possible for a cyclic reversible heat engine to convert all the absorbed heat into work without rejecting some heat at a lower temperature. This is against second law of thermodynamics Even a reversible (ideal) heat engine must reject a part of the absorbed heat to a sink at lower temperature. No cyclic engine can be 100% efficient.

6.10.How does behaviour of real gases differ from ideal gas at high pressure and low temperature? Identify the reasons behind these differences based on kinetic theory of gases.

At high pressure and low temperature, real gases deviate from ideal gas behavior due to:
1. Intermolecular forces: Unlike ideal gases, real gas molecules attract each other at close distances at low temperature, leading to deviate from ideal gas behaviour.
2. Finite molecular volume: Real gas molecules occupy space, so the free volume is less than predicted by the ideal gas law. As ideal gas molecules are assumed to point particles. At high pressure, molecules, are forced to closer together, and their size becomes significant, leading to deviation from ideal gas behaviour.
These effects, neglected in the kinetic theory of ideal gases, become significant under these conditions.

6.11.Show that area under P-V graph is equal to work done.

Graphical representation of work The work done by gas on piston can also be calculated by the area under the graph on PV-diagram as shown in Fig.
Area under PV graph = area of ABCD = length × width
W = ΔV × P
Area under PV graph

W = PΔV ----------(1)
∴ For constant pressure

W = PΔV ----------(2)
Comparing Eqs. (1) and (2)
Area under PV – graph = W (work done)
Hence area under PV graph shows the value of work done.

6.12.How is work done (i) by a gas (ii) on a gas? Calculate.

(i) Work done by the gas is taken as positive.
∴ W = Fd cosθ
Here
F = PA
d = Δy
θ = 0°
W = PA Δy cos0° = PA Δy ------------ (1) = PAΔy = PAV ∴ ΔΔy = Δv

(ii) Work done on the gas is taken as negative
W = Fd cosθ
Here F = PA, d = Δy and θ = 180°
W = PA Δy cos(180°) = PAA(– 1) = – PAΔ ∴ ΔV = ΔΔy
W = – PΔV ∴ ΔV = ΔΔy

SLO Based Additional Short Questions + Past papers Short Questions of Punjab Boards

Isothermal Process

Q1.A system undergoes from state P₁V₁ to state P₂V₂, as shown in Fig. What will be the change in internal energy?

There is no change in internal energy because the temperature remains constant. It is an isothermal process and during an isothermal process ∆U = 0.

Internal energy

Q2.Variation of volume by pressure is given in Fig. A gas is taken along the paths ABCDA, ABCA and A to A what will be the change in internal energy?

All processes represented in Fig. are cyclic, because in each case system returns to initial state therefore there is no change in internal energy.
Or ∆U = 0

Q3.Define the term internal energy.

The sum of all form of molecular energy (kinetic and potential) of substance is known as internal energy. It depends upon the temperature. It is state function. The change in internal energy depends upon its initial and final value and is independent of the path (process).

Heat engine

Q4.There is a huge reservoir of energy in ocean but we cannot use it. Explain.

To make use of energy we have to operate an engine which can only be done with the help of a source and a sink at different temperatures since there is no difference of temperature in sea, so we cannot use it.

Q5.A real heat engine is less efficient than Carnot engine. Explain.

In real heat engine there are forces of friction between the various parts of engine there are also heat losses. Therefore, part of output of engine is used up in doing work against these dissipative forces, which cause the decrease in its efficiency.

Entropy

Q6.Prove that maximum efficiency is always less than one or 100%.

We know the maximum efficiency of reversible engine is
η = (1 - Tc/Th) × 100
The heat energy takes heat energy from the source convert part of it in mechanical work and rest of it is rejected to sink. Therefore, the output of the engine is always less than input. The efficiency of heat engine is written is:
η = Output/Input × 100
η = ∆W/Q × 100
As ∆W < Q therefore the efficiency is less than 100%.

Heat engine

Q7.Is it possible to construct a heat engine that will not expel heat into the atmosphere?

For working of heat engine two bodies are required one at higher temperature that is call HTR and the other at lower temperature is called LTR or sink. Carnot engine take heat energy from HTR convert part of it in to mechanical work and the remaining part is rejected to sink. For the real heat engine atmosphere is sink hence during it operation it will expel heat to the atmosphere.

Entropy

Q8.An engine absorbs heat of 10 joule and reject 5 joule heat. What is the heat being used by the engine?

Let the heat engine takes heat energy Q₁ from the source and reject Q₂ to sink then heat energy used by engine to convert in mechanical work is Q₁ – Q₂ .
W = Q₁ – Q₂ = 10 – 5 = 5 Joule

Q9.Give four examples of a natural process that involve an increase in entropy.

The tide in the sea, the blowing of wind, the earthquakes, the radiation of the sun, all are examples due to which entropy is increased.

Q10.What is net change in the Entropy of a system when a Carnot cycle is completed?

We know that when Carnot cycle is completed, the net change in internal energy is zero. Hence the net change in entropy is zero.

Carnot engine

Q11.What is the effect on efficiency of Carnot engine, if temperature of the sink only is decreased?

When temperature of sink is decreased temperature difference increase this will increase the efficiency of heat engine.

First law of thermodynamics

Q12.What is the limitation of 1ˢᵗ law of thermodynamics which is covered by second law of thermodynamics?

1ˢᵗ Law of thermodynamics tells us that heat energy can be converted into equivalent amount of work. 2ⁿᵈ Law tell us the direction of flow of heat and also the amount of heat energy which can be converted into work. It tells us no heat engine is 100% efficient i.e. part of heat supplied must be rejected to a sink.

Q13.What is meant by metabolism? Apply 1ˢᵗ law of thermodynamic to explain it.

The process of transformation of energy that occurs within an organism is named as metabolism. From 1ˢᵗ law of thermodynamics, ∆u=Q-W
It tells us that the work done by moving body results in the decrease in internal energy of the body. This decrease is compensated by the energy provided due to the combustion of food.

Adiabatic expansion

Q14.Why does the temperature drop in adiabatic expansion?

When gas expands adiabatically; it is internal energy of the gas which is used up in doing work. Thus the internal energy of the system decreases. The temperature being directly related with internal energy drops.

Comparison between internal energy and gravitational P.E.

Q15.What is the similarity and difference between internal energy and gravitational P.E.?

The change in both type depends on their values at final state and independent of the path.
The gravitational P.E has dependence on the gravitational force, whereas internal energy has no such dependence on the field of force.

Entropy

Q16.Show that change in entropy is always positive.

All natural processes are irreversible in which entropy of system increases. Therefore, change is positive.

Constructed response Questions

6.1.Explain how thermodynamics relates to the concept of energy conservation.

First law of thermodynamics expresses the law of conservation of energy by affirming that energy is conserved quantity in isolated systems. It provides a framework to understand how energy transferred and transformed within systems without violating the fundamental principle that energy cannot be created nor destroyed. This alignment underscores the broader applicability and importance of first law in understanding the behaviour of energy in the universe.

6.2.Explain how thermodynamics applies to biological systems, such as humanbody.

Human Metabolism
Human Metabolism also provides an example of energy conservation. Human beings and other animals do work when they walk, run, or move. Work requires energy. Energy is also needed for growth to make new cells and to replace old cells that have died. Energy transforming processes that occur within an organism are named as metabolism. We can apply the first law of thermodynamics.
∆U = Q – W
To an organism of the human body. Work done will result in the decrease in internal energy of the body. Consequently, the body temperature or in other words internal energy is maintained by the food we eat.

6.3.A gas is expanding adiabatically. Explain what happens to temperature and pressure of the gas.

In adiabatic expansion, no heat is exchanged with the surrounding. As the gas expands, it does work on the surroundings, due to which internal energy decreases. This leads to decrease in temperature and pressure of the gas.

6.4.A coffee cup is left on a table, and overtime coffee cup cools down. Explain thermodynamics processes occurring during this process.

When a coffee cup is left on the table, it loses heat to the surroundings through convection, conduction and radiation according to second law of thermodynamics, heat flows from coffee to cooler environment (surrounding) until thermal equilibrium is reached.

6.5.How we can explain different weather patterns through thermodynamically processes like wind, rain, etc.

Weather patterns like wind, rain and storms are results of thermodynamic processes the sun heats the earth unevenly, causing Temperature and pressure differences in the atmosphere. These differences drive winds, cloud formation (condensation of water vapour) and precipitation (release of latent heat), all governed by the laws of thermodynamics.

Comprehensives Questions

6.1.What are the postulates of kinetic theory of gases? Derive a relation for ideal gas equation in the form PV=NkT from general gas equation.

See Q.1 and Q.4 of theory.

6.2.State and explain various gas laws.

See Q.5 of theory.

6.3.Explain first law of thermodynamics in detail. Give an example in support of your explanation. Give its two applications.

See Q.9 of theory.

6.4.What is a refrigerator? Explain its working. Derive an expression for its co-efficient of performance.

See Q.15 of theory.

6.5.What is Carnot engine? Describe Carnot cycle. State Carnot theorem and derive an expression for efficiency of Carnot engine.

See Q.14 of theory.

6.6.Define and explain the term "Entropy".

See Q.16 of theory.